The Stacks project

Proof. Let $R$ be a principal ideal domain. Let $g = a/b$ be an element of the fraction field of $R$ integral over $R$. Because $R$ is a principal ideal domain we may divide out a common factor of $a$ and $b$ and assume $(a, b) = R$. In this case, any equation $(a/b)^ n + r_{n-1} (a/b)^{n-1} + \ldots + r_0 = 0$ with $r_ i \in R$ would imply $a^ n \in (b)$. This contradicts $(a, b) = R$ unless $b$ is a unit in $R$. $\square$

Comments (2)

Comment #5508 by Manuel Hoff on

The same proof shows that in fact any factorial ring is normal. One just has to replace with the (weaker) condition that and are coprime in the sense that their greatest common divisor is . This is enough to make the conclusion.

There are also:

  • 3 comment(s) on Section 10.37: Normal rings

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00GZ. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 10.37.6 as pdf